Question

TRIANGLES
5. In Fig. 6.20, DE | OQ and DF OR Sbow that
EF OR​

Answer 1

Step-by-step explanation:

In triangle POQ ,DE parallel OQ

PE/EQ = PD/DO ( Basic promotional theorem)

Now, In triangle POR ,DF parallel OR

PF/FR =PD/ DO

From in eq. 1 and 2 we have

PE/EQ = PF/ FR

therefore EF parallel QR (convers of basic proportional theorem)

Answer 2

Answer:

HELLO MATE ,

EF//QR [CONVERSE OF BPT]

Step-by-step explanation:

GIVEN;

DE//OQ

DF//OR

PROVE;

EF//QR

SOLUTION;

IN ΔPOQ ,  DE//OQ

       $\frac{PE}{EQ} =\frac{PD}{DO} ------[BPT]--1$

IN ΔPOR ,  DF//OR

      $\frac{PE}{FR} =\frac{PF}{FR} ------[BPT]--2$

From equation 1 and 2

      $\frac{PE}{EQ} =\frac{PF}{FR}$

IN ΔPQR  ,  EF//QR

      $\frac{FE}{EQ} =\frac{PF}{FR} ------[proved]$

THEREFORE, EF//QR [converse of BPT]

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